3.410 \(\int \frac{(a+i a \tan (c+d x))^{5/2}}{\sqrt{e \sec (c+d x)}} \, dx\)

Optimal. Leaf size=563 \[ \frac{5 i a^{7/2} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a^{7/2} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}-\frac{5 i a^{7/2} \sec (c+d x) \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{5 i a^{7/2} \sec (c+d x) \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}} \]

[Out]

((5*I)*a^(7/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c +
d*x])/(Sqrt[2]*d*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((5*I)*a^(7/2)*ArcTan[1 + (
Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(Sqrt[2]*d*Sqrt[e]*S
qrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/2)*a^(7/2)*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sq
rt[a - I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d
*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (((5*I)/2)*a^(7/2)*Log[a + (Sqrt[2]*Sqrt[a]*
Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/
(Sqrt[2]*d*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((10*I)*a^2*Sqrt[a + I*a*Tan[c + d
*x]])/(d*Sqrt[e*Sec[c + d*x]]) + (I*a*(a + I*a*Tan[c + d*x])^(3/2))/(d*Sqrt[e*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.575009, antiderivative size = 563, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3498, 3496, 3499, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac{5 i a^{7/2} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a^{7/2} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}-\frac{5 i a^{7/2} \sec (c+d x) \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{5 i a^{7/2} \sec (c+d x) \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2)/Sqrt[e*Sec[c + d*x]],x]

[Out]

((5*I)*a^(7/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c +
d*x])/(Sqrt[2]*d*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((5*I)*a^(7/2)*ArcTan[1 + (
Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(Sqrt[2]*d*Sqrt[e]*S
qrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/2)*a^(7/2)*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sq
rt[a - I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d
*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (((5*I)/2)*a^(7/2)*Log[a + (Sqrt[2]*Sqrt[a]*
Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/
(Sqrt[2]*d*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((10*I)*a^2*Sqrt[a + I*a*Tan[c + d
*x]])/(d*Sqrt[e*Sec[c + d*x]]) + (I*a*(a + I*a*Tan[c + d*x])^(3/2))/(d*Sqrt[e*Sec[c + d*x]])

Rule 3498

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3499

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(d*Sec
[e + f*x])/(Sqrt[a - b*Tan[e + f*x]]*Sqrt[a + b*Tan[e + f*x]]), Int[Sqrt[d*Sec[e + f*x]]*Sqrt[a - b*Tan[e + f*
x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rule 3495

Int[Sqrt[(d_.)*sec[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-4*b*d^
2)/f, Subst[Int[x^2/(a^2 + d^2*x^4), x], x, Sqrt[a + b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]]], x] /; FreeQ[{a, b,
d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^{5/2}}{\sqrt{e \sec (c+d x)}} \, dx &=\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}}+\frac{1}{2} (5 a) \int \frac{(a+i a \tan (c+d x))^{3/2}}{\sqrt{e \sec (c+d x)}} \, dx\\ &=-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}}-\frac{\left (5 a^3\right ) \int \frac{(e \sec (c+d x))^{3/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{2 e^2}\\ &=-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}}-\frac{\left (5 a^3 \sec (c+d x)\right ) \int \sqrt{e \sec (c+d x)} \sqrt{a-i a \tan (c+d x)} \, dx}{2 e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}}-\frac{\left (10 i a^4 e \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{a^2+e^2 x^4} \, dx,x,\frac{\sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}\right )}{d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}}+\frac{\left (5 i a^4 \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{a-e x^2}{a^2+e^2 x^4} \, dx,x,\frac{\sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}\right )}{d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{\left (5 i a^4 \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{a+e x^2}{a^2+e^2 x^4} \, dx,x,\frac{\sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}\right )}{d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}}-\frac{\left (5 i a^4 \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{a}{e}-\frac{\sqrt{2} \sqrt{a} x}{\sqrt{e}}+x^2} \, dx,x,\frac{\sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}\right )}{2 d e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{\left (5 i a^4 \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{a}{e}+\frac{\sqrt{2} \sqrt{a} x}{\sqrt{e}}+x^2} \, dx,x,\frac{\sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}\right )}{2 d e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{\left (5 i a^{7/2} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{a}}{\sqrt{e}}+2 x}{-\frac{a}{e}-\frac{\sqrt{2} \sqrt{a} x}{\sqrt{e}}-x^2} \, dx,x,\frac{\sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}\right )}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{\left (5 i a^{7/2} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{a}}{\sqrt{e}}-2 x}{-\frac{a}{e}+\frac{\sqrt{2} \sqrt{a} x}{\sqrt{e}}-x^2} \, dx,x,\frac{\sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}\right )}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=-\frac{5 i a^{7/2} \log \left (a-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{5 i a^{7/2} \log \left (a+\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}}-\frac{\left (5 i a^{7/2} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (5 i a^{7/2} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{5 i a^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right ) \sec (c+d x)}{\sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a^{7/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right ) \sec (c+d x)}{\sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a^{7/2} \log \left (a-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{5 i a^{7/2} \log \left (a+\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{2 \sqrt{2} d \sqrt{e} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{10 i a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{e \sec (c+d x)}}+\frac{i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt{e \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.93167, size = 367, normalized size = 0.65 \[ -\frac{a^2 \sqrt{a+i a \tan (c+d x)} \left (\sqrt{-\sin (c)+i \cos (c)+1} \left (\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} (\tan (c+d x)+9 i)-5 i \sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )\right )-5 i \sqrt{-\sin (c)-i \cos (c)+1} \sqrt{-\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)+1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)+1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )\right )}{d \sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\sin (c)+i \cos (c)+1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Sqrt[e*Sec[c + d*x]],x]

[Out]

-((a^2*Sqrt[a + I*a*Tan[c + d*x]]*((-5*I)*ArcTan[(Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[1
- I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[1 - I*Cos[c] - Sin[c]]*Sqrt[-1 + I*Cos[c] - Sin[c]]*Sqrt[I
+ Tan[(d*x)/2]] + Sqrt[1 + I*Cos[c] - Sin[c]]*((-5*I)*ArcTan[(Sqrt[-1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/
2]])/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]
] + Sqrt[-1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]]*(9*I + Tan[c + d*x]))))/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[-
1 + I*Cos[c] - Sin[c]]*Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]]))

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Maple [A]  time = 0.345, size = 347, normalized size = 0.6 \begin{align*}{\frac{{a}^{2}}{2\,d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) \cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 5\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ){\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-5\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}{\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) -5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ){\it Artanh} \left ( 1/2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}{\it Artanh} \left ( 1/2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) -16\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+18\,i\cos \left ( dx+c \right ) +16\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -2\,i+2\,\sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x)

[Out]

1/2/d*a^2*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(5*I*cos(d*x+c)*sin(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)
+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*(1/(cos(d*x+c)+1))^(1/2)-5*I*cos(d*x+c)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1
/2)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))-5*cos(d*x+c)*sin(d*x+c)*arctanh(1/2*(1/(co
s(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*(1/(cos(d*x+c)+1))^(1/2)-5*cos(d*x+c)*sin(d*x+c)*(1/(cos(d*x+c)+
1))^(1/2)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))-16*I*cos(d*x+c)^2+18*I*cos(d*x+c)+16
*cos(d*x+c)*sin(d*x+c)-2*I+2*sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)/(e/cos(d*x+c))^(1/2)

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Maxima [B]  time = 2.47983, size = 2720, normalized size = 4.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

((80*sqrt(2)*a^2*cos(2*d*x + 2*c) + 80*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 80*sqrt(2)*a^2)*arctan2(sqrt(2)*cos(1/
4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
)) + 1) + (80*sqrt(2)*a^2*cos(2*d*x + 2*c) + 80*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 80*sqrt(2)*a^2)*arctan2(sqrt(
2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c))) + 1) + (80*sqrt(2)*a^2*cos(2*d*x + 2*c) + 80*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 80*sqrt(2)*a^2)*arc
tan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c))) + 1) + (80*sqrt(2)*a^2*cos(2*d*x + 2*c) + 80*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 80*sqrt(2)
*a^2)*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1, -sqrt(2)*sin(1/4*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (80*I*sqrt(2)*a^2*cos(2*d*x + 2*c) - 80*sqrt(2)*a^2*sin(2*d*x + 2*c) +
80*I*sqrt(2)*a^2)*arctan2(sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c))), sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (-80*I*sqrt(2)*a^2*cos(2*d*x + 2*c) + 80*sqrt(2)*a^2*sin(2*d*x +
2*c) - 80*I*sqrt(2)*a^2)*arctan2(-sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arct
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), -sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - (512*a^2*cos(2*d*x + 2*c) + 512*I*a^2*sin(2*d*x + 2*c)
+ 640*a^2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (40*sqrt(2)*a^2*cos(2*d*x + 2*c) + 40*I*sqr
t(2)*a^2*sin(2*d*x + 2*c) + 40*sqrt(2)*a^2)*log(2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))
*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c))) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c)))^2 + 2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - (40*sqrt(2)*a^2*cos(2*d*x + 2*c) + 40*I*sqrt(2)*a^2*sin(2*d*x + 2*c)
+ 40*sqrt(2)*a^2)*log(-2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d
*x + 2*c), cos(2*d*x + 2*c))) - 2*(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1)*cos(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*cos(1
/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2
*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c))) + 1) + (40*I*sqrt(2)*a^2*cos(2*d*x + 2*c) - 40*sqrt(2)*a^2*sin(2*d*x + 2*c) + 40*I*sqrt(2)*a^2)*log
(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c))) + 2) + (-40*I*sqrt(2)*a^2*cos(2*d*x + 2*c) + 40*sqrt(2)*a^2*sin(2*d*x + 2*c) - 40*I
*sqrt(2)*a^2)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2*sqrt(2)*sin(1/4*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + (40*I*sqrt(2)*a^2*cos(2*d*x + 2*c) - 40*sqrt(2)*a^2*sin(2*d
*x + 2*c) + 40*I*sqrt(2)*a^2)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*s
qrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + (-40*I*sqrt(2)*a^2*cos(2*d*x + 2*c) + 40*sq
rt(2)*a^2*sin(2*d*x + 2*c) - 40*I*sqrt(2)*a^2)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 +
2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c))) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + (-512*I*a^2*cos(2*d*x + 2*
c) + 512*a^2*sin(2*d*x + 2*c) - 640*I*a^2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sqrt(a)*sqrt(
e)/((-64*I*e*cos(2*d*x + 2*c) + 64*e*sin(2*d*x + 2*c) - 64*I*e)*d)

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Fricas [A]  time = 2.35764, size = 1615, normalized size = 2.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(25*I*a^5/(d^2*e))*d*e*e^(I*d*x + I*c)*log(2/5*(sqrt(25*I*a^5/(d^2*e))*d*e*e^(2*I*d*x + 2*I*c) + 5*(a
^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x
+ 3/2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) - sqrt(25*I*a^5/(d^2*e))*d*e*e^(I*d*x + I*c)*log(-2/5*(sqrt(25*I*a^5/(d
^2*e))*d*e*e^(2*I*d*x + 2*I*c) - 5*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e
^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) - sqrt(-25*I*a^5/(d^2*e))*d*e*e^(I
*d*x + I*c)*log(2/5*(sqrt(-25*I*a^5/(d^2*e))*d*e*e^(2*I*d*x + 2*I*c) + 5*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(
a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c))*e^(-2*I*d*x - 2*I*c)/a
^2) + sqrt(-25*I*a^5/(d^2*e))*d*e*e^(I*d*x + I*c)*log(-2/5*(sqrt(-25*I*a^5/(d^2*e))*d*e*e^(2*I*d*x + 2*I*c) -
5*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I
*d*x + 3/2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) + 2*(-8*I*a^2*e^(2*I*d*x + 2*I*c) - 10*I*a^2)*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c))*e^(-I*d*x - I*c)/(d*e)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(5/2)/(e*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sqrt{e \sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)/sqrt(e*sec(d*x + c)), x)